Projectivities of free products

R ^ENDICONTI

del

S ^EMINARIO M ^ATEMATICO

della

U NIVERSITÀ DI P ^ADOVA

C HARLES S. H OLMES Projectivities of free products

Rendiconti del Seminario Matematico della Università di Padova, tome 42 (1969), p. 341-387

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by

^{CHARLES S. HOLMES}

*)

RESUME. A projectivity ^{of a} group G ^{onto a} group H is ^an isomorphic mapping of the subgroup ^lattice L(G) ^{of G onto the} subgroup ^lattice L(H) ^{of H.}

Considerable attention has been devoted to the following questions: ^under what conditions on G is _any projectivity ^{of G induced} by ^some isomorphism

of G, ^{and when must G} be determined by ^its lattice of subgroups ^{in the sense}

that any group H with L(H) isomorphic ^to L(G) ^must itself be isomorphic

to G? Theorems 4 and 6 of this _paper are partial ^{answers to these} questions for the class of _groups which have a non-trivial decomposition ^{as a free} product ^with amalgamated subgroup.

Let G ⁼ (A, B # C) ^{be a free} product ^{of its} subgroups ^{A and B}

with amalgamated subgroup C = A tl B. Suppose [A : C] &#x3E; 2 ^or [B : C] &#x3E; 2.

Theorem 4. If C is normal in G, ^{then G is} determined by its lattice of sub- groups. Theorem 6. If C is the center of G, then every projectivity ^{of G}

is induced by ^a unique group isomorphism. Corollary. Any projectivity ^of

a free product ^A*B (A, B ~ 1) ^is induced by ^a unique isomorphism. ^The proofs depend upon the fact that A’~B(A, ^{contains a} non-cyclic ^free

group unless A and B ^are both of order two. The methods used ^are primarily

extensions and refinements of those used by ^{E. L. Sadovskii} (Mat. ^Sbomik 21 (63) (1947), 63-82).

Introduction.

The collection of all

subgroups

^{of a} group G forms a lattice

L(G);

here the lattice meet, H ⁿ

K,

of two

subgroups

^{H and K is their inter-}

section H n K,

^{while the lattice}

join,

^{H v}

K,

^{is the}

subgroup generated by

the union H U K.

Following

^Suzuki

[4]

^{we define a}

projectivity

*) Indirizzo dell’A.: Depart ^of Math., ^Miami University, Oxford, ^Ohio 45056, U.S.A.

from a group G onto a group H to be an

isomorphic mapping

^{of the}

subgroup

^lattice

^L(G)

^{onto the}

subgroup

^lattice

^L(H).

^For the definitions of

concepts

^discussed

informally

ⁱⁿ the introduction the reader is refer- red to the

section,

Definitions and

Notations,

^{of this} paper.

Every isomorphism

^{from a} group G onto a group H

clearly

^induces

a

projectivity

^{from G onto} H. Thus if G and H are

isomorphic, ^L(G)

and

L(H)

^are

isomorphic.

Considerable attention has been devoted to the converse

questions:

under what conditions on G is _any

projectivity

of G induced

by

^some

isomorphism

^of

^G,

^{and when must G and H be}

isomorphic

^if

L(G)

^and

L(H)

^are

isomorphic?

It is _easy to see that

non-isomorphic

groups may have

isomorphic

lattices of

subgroups;

^{hence a} group need not be determined

by

^its

lattice of

subgroups

^{and a}

projectivity

^{need not} be induced

by

^{an iso-}

morphism.

^The

simplest examples

^{are the} groups of

prime order,

^{all of} which have the two element lattice as

subgroup

^{lattice. There are less}

trivial

examples

^of groups which are not determined

by

^their

subgroup

lattices. On the other hand several classes of groups have been shown

to consist of _groups which are determined

by

their lattices.

Sadovsky,

^for

^instance,

has shown the

following.

^{If G=A*B is}

a free

product,

^without

amalgamation,

of non-trivial _groups A and

B,

then G is determined

by

^its lattice of

subgroups ^L(G).

^{He also} specu- lated that _any

projectivity

p of such a free

product

^G=A*B is induced

by

^an

isomorphism,

but left the

question

open. He made several con-

tributions to the solution of this

problem,

^{two of which are} useful here.

First he showed that if an

inducing isomorphism

^{exists for a}

projecti- vity

^{of a free}

product,

^{then it must be}

unique. Sadovsky

^{also showed}

that _any

projectivity

^{of a free}

product

^G=A*B where A and B are both of order two is induced

by

^an

isomorphism.

^{It is an immediate conse-}

quence of Theorem 6 in this _paper that _any

projectivity

^{of a free}

product

G = A*B where A or B is of order greater ^{than two is} induced

by

^an

isomorphism

^of

^G; again

^{A and B are} non-trivial.

Combining

^these

results we have the

following

^result.

Any projectivity

^{of a free}

product

G=A*B with non-trivial factors A and B is induced

by

^a

unique

^iso-

morphism.

In this _paper we

partially

^{answer the same}

questions

^{for the more}

general

^{class of free}

products G = A~ B

^with

amalgamated subgroup

C=AnB,

^{where C is a} proper

subgroup

^{of both A and B. For the}

definition and

elementary properties

^{of free}

products

^{and free groups}

we refer the reader to the

Theory

^of

Groups [2] by

^{Kurosh. In Theorem}

4 it is shown that _{any group}

G=A]B

^is determined

by

^its lattice of

subgroups L(G),

if C is a proper normal

subgroup

^{of A and}

^B,

^{and the}

index of C in

A, [ A : C],

^{or the index of C in}

B, [B : C],

^is

strictly

greater ^{than two. If in} addition C is the center of

G,

every

projectivity

p of G is induced

by

^a

unique isomorphism

^{of G}

^{(Theorem 6).}

These results are

certainly

^{not the best}

possible.

^{It is not difficult}

to see that the

arguments given

^below

permit

^{us to draw the same con-}

clusions with the conditions on C weakened in various

significant

^ways.

However,

^{we have not stated} any such extensions of our main

result, partly

^{because we have not been able to} prove any that does not involve rather

complicated

and artificial

conditions,

and

partly

^{because we have}

no evidence that the conclusions do not hold

quite generally

^{with the}

conditions on C

(except

^for C proper in A and

B) simply dropped.

^It

seems evident that our methods could ^not

yield

^{such a} result without considerable revision and extension.

The methods used here are based _upon the

following

^{two facts.}

The first fact is that _any

projectivity

p from a group G ^{onto a} group H is

fully

determined

by

its action on the

cyclic subgroups [g]

gene- rated

by

the elements _g of

G,

^since every

subgroup

^K of G is the

join

of the

cyclic subgroups

^{contained in K.} The second fact

(due

^{to Baer}

L 1 ] )

is that the

image

^{of a}

cyclic subgroup

^{must be}

cyclic.

^That

^is,

^for

every element _g of G there is an element h

(not necessarily unique)

such that

p [ g ] _ [ h ~ . Combining

^{the two results we see that the}

image

^{of the}

subgroup [gi]

^v under the

projectivity

p ^must be

p([gil

^v

[g2] )

^#

Ehil

^v

[h2]

^where

p[gi] = ^[hi]

^and

p[g2] = [h2] .

^In

this connection it is

simple

^{to observe that}

hence that

where

p[gig2] = [h].

^{This use of the}

join preserving property

^{of the}

projectivity

^{is one of the rare} instances in which lattice

theory

^enters

into our arguments.

For the moment let G = A*B be a free

product,

^without

amalgama- tion,

of non-trivial

subgroups

^{A and}

B,

^{and let} p be a

projectivity

^from

some group G’ onto G. Let A’ and B’ be the

subgroups

^{of G’ such}

that

pA’ = A

^and

pB’=B. Sadovsky

found that he could

pick

^elements

v’ in G’ and v in G with

p [ v’ ] _ [ v ]

^{in such a way} that the relation

was an

isomorphism

^{from A’ onto A. This}

isomorphism

^{in turn induced}

the restriction of _p on A’. In a similar _way

isomorphisms

^{were determined}

from B’ ^onto B which also induced the

projectivity

^{on B’.}

Sadovsky

then showed that G’ was a free

product

of A’ and

B’,

hence that G and G’ were

isomorphic.

^{He left} open the

problem

^of

extending

^the

isomorphisms

^{on the} components ^{to an}

isomorphism

^{on G’}

inducing

p.

Now let _p be ^a

projectivity

^{of G’ onto where C is a} proper normal

subgroup

^{of A and B, and}

^{[A : C]}

^or

^{[B : C]}

^is

strictly

greater than two.

Again

let A’ and B’ be the

subgroups

of G’ such that

pA’=A,

and

pB’=B.

^{In Theorem 3 of this} paper it is shown that for _any element

g’

in G’ there is an

isomorphism

^from

[g’]

^{v C’ onto}

p[g’]

^{v C.}

Shortly

thereafter it is shown that there are

isomorphisms f

^{1 and}

f 2

^which map A’ onto A and B’ onto

B, respectively,

^{such that}

f

^{1 and}

f 2

agree ^on C’.

We define

f

^{to be the} map

pieced together

^{from the}

isomorphisms

^bet-

ween

[g’]

^{v C2 and}

p[g’]

^{v C. In Theorem 4 it is} shown that G’ is

isomorphic

^{to G}

by showing

^that

f

^is the natural extension of

11

^{1 and}

f 2

^to all of G’.

1. Def initions and Notation.

If gi , ..., gn ^are elements of the _group

G,

^we shall write

[gi , ..., gn]

for the

subgroup

^{of G}

generated by

gli , ..., gn . We shall also write

[g, , ...,

gn , ^...,

Um ]

^{for the}

subgroup generated by

the elements gi

together

^{with the subsets}

U;

of G. Thus if G1 and G2 are two

subgroups

of

G,

^then

[ G1 , G2] = G1 v G2 . Throughout

^the paper ^we let 1 stand for the

identity subgroup

^or the first natural

number,

^as

appropriate.

If H is a

subgroup

^{of the} group

G,

^we write H G. If X and Y are subsets of a set

S,

^{we write X-Y for the}

complement

^{of Y in X.}

We record here the fundamental definitions of this

subject

^and

presume that the introduction

provides ample

discussion of these

topics.

DEFINITION. Let G and H be _groups, and let

L(G)

^and

L(H)

^be the lattices of

subgroups

^{of G and}

^H, respectively.

^A

mapping

p from

L(G)

^into

L(H)

^{is a}

projectivity

^{of G onto H if and}

only

^if p is ^{a one} to one

mapping

^of

^L(G)

^onto

^L(H)

^{and for any two}

subgroups

^{G1 and}

G2 ⁱⁿ

G, p(G1 n G2)=pGi fl pG2

^and

p[ G1 , G2] _ [ pGl , pG2].

^Two

subgroup

^lattices

^L(G)

^and

^L(H)

^{are said to be}

isomorphic

^{if and}

only

if there is ^a

projectivity

^{from G onto H.}

DEFINITION. A group G is determined

by L(G),

^its lattice of sub- groups, if and

only

^if any group H with

L(H) isomorphic

^to

^L(G)

^must

itself be

isomorphic

^{to G. A}

projectivity

p from a group G ^{onto a} group H is said to be induced

by

^an

isomorphism,

^{if there is an}

isomorphism o from

^{G onto H such that for} any

subgroup

^{K in}

^G, kEK }.

We shall use the notation

G = A~ B

^{to mean that G is the free}

product

^{of its}

subgroups

^{A and B with}

amalgamated subgroup

^{C = AnB.}

When C is the

identity subgroup

^we shall write

simply

^{G=A*B and}

call G the free

product

^{of A and}

B,

^without

amalgamation.

^{In fact we}

shall never use this notation except under the

assumption

^{that C is}

properly

^{contained in both A and}

^B;

^that

is,

^{we assume} that A - C and B-C are non-empty.

DEFINITION. Let

G = A~ B.

^Then any

expression

gi ^... gn with gi

alternately

ⁱⁿ A - C and B - C and with n a

positive integer

^{is said}

to be a word in G.

In a free

product G = A* c B

any word _gl ^... _gn considered as a

product

^{in G is an} element of the

complement

^{of C in}

^G,

G - C. Con-

versely

every element in G - C can be

expressed

^{as a word in G. For}

an element w in G - C this

representation

^... gn ^{as a} word in G is not in

general unique,

^{but vv does determine}

uniquely

^the

length

n and the double cosets

Cg1C,

^...,

CgnC.

In order ^to see this let _gi ^... _gm and

hi ... hn

^{be two} words which represent ^{the same} element in

G - C,

^{that is} gl ^...

gm=h1

^{... hn .}

Then

his

^{1 ... ...}

gm =1,

^{where 1} stands for the

identity

^element

in G. Now

hi 1 gi

^{must be in}

C,

^{for otherwise the}

expression

^{on the left}

is a word or can be written as one. That is if hl ^and gl ^are from the

same group and

h1181

^{is not in}

C,

^{then and} gi ^are all from

the same set A - C or B - C and

is a word as written. In a

similary

way if

hi

^and _gi ^are from different groups then the

expression

is a word as written. A

simple

induction shows that ^... _gi is in

C,

where i is less than or

equal

^{to m and n. Hence} that m = n and that hi ^and _gi ^are both in the same set A - C or B - C. Moreover the double cosets

ChiC

^and

CgiC

^are

equal

^{for each}

i =1, ...,

m, ^or

simply

when C is normal. When C is the

identity subgroup

^{of G or}

in other words when G = A*B is a free

product

^without

amalgamation,

every

non-identity

^{element in G has a}

unique representation

^{as a word}

in G.

DEFINITION. Let

G = A~ B. Suppose

^that g is an element of the

complement

^{of C in G and that} g==g] ^... gn is _any

representation

^for

g ^{as a} word in G. If _gl is in A - C or

B - C,

^we say that _g

begins

^with

an element of A or

B, respectively.

^If gn is in A - C or

B - C,

^we say that _g ends with an element of A or

B, respectively.

^{We also} say that the

length I g I

^of g is _n, and take the

length

^of any element in C to be zero.

The

following

^{results are}

simple

consequences of these definitions.

Let _g and h be _any two elements in the

complement

^{of C in}

G=A*B.

Then

2)

and

only

^if

g-I

^{and b both}

begin

^with

elements from the same group A or B.

3)

^If

I g I

^and

gh

^{is not an} element of

C,

^then

gh

^and g

begin

^{with elements of the same} group while

gh

^{and h end with elements}

from the same group.

Now let A : A

designate

^{the set of elements in G-C}

beginning

and

ending

^{with elements of}

^A,

^{and A : B the set of elements}

beginning

with elements of A and

ending

with elements of B. Also let B : A and B : B be defined

similarly.

^{It is} easy ^{to see} that the set G is the

disjoint

^{union of}

^C,

^{A :}

^A,

^{A :}

^B,

^{B :}

^A,

and B : B.

2.

Elementary Propositions.

In this section we isolate some rather technical

propositions

^which

are useful in the

following

^section.

Throughout

^the paper ^we shall reserve

the word

proposition

for facts from group

theory

^{and the word lemma}

and theorem for the results about

projectivities

^of groups.

PROPOSITION 1. Let G be a group, N a normal

subgroup

^{and V}

a non-trivial

cyclic subgroup

^{of G. Let U and W be}

cyclic subgroups

of G.

Suppose

Then for _any generator ^{w of W there is a}

unique

^{u in N and there is}

a

unique v

^{in V such that}

Moreover

[ v ] = V

^and

[ u, V].

PROOF.

Suppose

^{w is some} generator of W. From

W [ N, V]

with N normal in G it follows that there are elements u in N and v in V such that ^{w = uv.} If also w = uivi with ul in N and vi in

V,

^then and

VIV-1

are

equal

^{and therefore}

equal

^{to the}

identity

^element

in G since V f1 N== 1. Hence u and v are

unique

^{in N and V}

respectively.

From

[ U, W] = [U, V]

^{and U:5 N it follows that}

Hence

for otherwise

Thus

[ v] = V.

^{But then}

and the

proposition

^is

proved.

COROLLARY. If in

Proposition

^{1 the}

subgroup

^{N is in the center}

of

G,

^then

[ u ] = U.

PROOF. Since N is in the ^center of G the

subgroup [ U, V]

^{is the}

direct

product

^{of U and V.} Hence u is an element of U. Then from

[ u, V ] _ [ U, V ]

the conclusion follows.

It _{is easy} to see that in

Proposition

^{1 we could have concluded}

that there are

unique

^{elements u in U} and v in V such that vv = vu.

PROPOSITION. 2. Let G be _{any group} and

U,

^{V and W} non-trivial

cyclic subgroups

^{of G.}

Suppose

^that

Then there is a

unique

^u in U and there is a

unique

^{v in V such that}

Moreover

[ u ] = U

^and

[ v ] = V.

PROOF. We assume that W is not in U or

V,

^{for otherwise} U*V = U ^or U*V = V. Let w be a generator of ^W.

Suppose

^{that U.}

Then

and all the elements of

[V, W]

^which

begin

and end with elements of U have

length

^{three or more. That}

is,

^{U and}

imply

the contradiction

Thus U.

Similarly w 0 V :

^V.

Now w e U : V or w E V :

U,

^{and w-1} is in the other. We _may suppose that we U : V. Let

If

1 C n,

^{then U but} and we

again

^{have the} contradiction

Thus

Since U*V is a free

product,

^without

amalgamation,

^{there are no other}

elements

u2 E U,

v2 E V such that Furthermore W is infinite

cyclic

^{and its}

only

^other generator ^{is U. So ui and}

vi ^are the

only

^two elements of G such that

We write

simply w=uv.

^The fact that

[ u ] = U

^and

[ v ] = V

^fol-

lows

immediately

from the relations

Again

^{it is} easy ^{to see} that in

Proposition

^{2 we} could have con-

cluded that there exist

unique

^{elements u in U and v in V such that}

W= [vu].

PROPOSITION 3. Let G=

[t]*[v]*[x,

y,

z]

^{where t and v are not}

of order two.

Suppose

^that

where

j, k, I,

m, n = ± 1. If or m,

n =1,

^then

PROOF. Let

T = [ t ] , V = [ v ]

^and

X = [ x,

y,

z].

Let w be the element

given by

^the

equated expressions.

^The

expression

^{on the}

right

shows that 1 ends with an element of V.

Hence ;==1

^and

Suppose

first that m, n=1. Then k= -1

implies

^that

begins

^with

x -1

or t-I rather than t as it must. Hence k =1.

Similarly

^{1= 1.}

Suppose

^{now that} y is non-trivial. Since w ends with an element of

V, l =1

^{and Thus}

Let u be the element

given by

^these

expressions.

Then either u does

or does not end with an element of X. If u does end with an element of

X,

^{then m =1 and we are done. If u does not end with an element} of

X,

then k =1 and x is the inverse of y. Now

m =1,

for otherwise t is its own inverse.

3. Functions Induced

by

^a

Projectivity.

Let _p be a

projectivity

^{from a group G’ onto a group G. Let v’}

be an element of G’ and v an element of G such that

p [ v’ ] _ [ v ] .

^Let

U’ be a

subgroup

^{of G’ and U the}

image

^{of U’ under} p.

DEFINITION.

Suppose

^{that for} every u’ in U’ there is a

unique

u in U such that

p[u’v’] = [uv] (or p [ v’u’ ] _ [ vu ] ) .

Then the function

f

^defined

by f (u’) = u

^{for all} elements u’ in U’ is said to be the function on U’ induced

by

^the

projectivity

p and the

pair (v’, v)

with v’ and v on the

right (left).

For a

given f unction f

from U’ into U ^we write

f = f { .; v’, v) ( f = f (v’,

^v;

.))

^{if and}

only

^if

f

^{is the function induced}

by

^the

projec- tivity

p and the

pair (v’, v)

^with

(v’, v)

^{on the}

right (left).

^{Thus for}

any element u’ in

U’,

We also _say that a function

f

from U’ into U is induced

by

^the

projec- tivity

p if and

only

^{if there is a}

pair

of elements v’ in G’ and v in G such that

f = f (.; v’, v)

^or

f = f (v’,

^v;

.).

^{Because there is}

usually only

one

projectivity

p under

discussion,

^{we sometimes}

simply

^{say that the}

pair (v’, v)

induces the functions

f(v’,

^v;

^.)

^and

f(.; v’, v).

In this section we shall show that functions

f = f(.; v’, v)

^{do exist}

for certain

subgroups

^{U’ and U and}

pairs

^{v’ and v.} In fact if U’ and U are the

identity subgroups

^{of G’ and}

^G, respectively,

^{and if}

f

^is

the function which _maps U’ onto U in the

only possible

^{way then}

f=f(.; v’, v) = f (v’,

^v;

.).

^{In the more}

general

situation in which U’

and U are not trivial _groups there _may be non-trivial elements u in U such that

p [ v’ ] _ [ uv ] .

^For

example

suppose that v has order twelve in G and that If v’ is _any element of G such that

p[v’]

⁼

[ v ] ,

^then

p [ v’ ]

⁼

[ ( v4)v ] .

That is the element v4 also

corresponds

to the

identity

^{of G’.} Thus in the

following

^{work we} shall almost

always

assume that

[ v ]

^{meets U in the}

identity subgroup.

For the remainder of this section let _p be a

projectivity

^{of some}

group G’ onto a group G. If H is _any

subgroup

^{of G we write H’ for}

the

subgroup

^{of G’} such that

pH’ = H.

LEMMA 1. Let N be a normal

subgroup

of G and V an infinite

cyclic subgroup

^{of G} such that V fl N=1.

Suppose

^{v’ and v are} gene- rators of V’ and

V, respectively.

Then for each u’ in

N’,

^{there is a}

unique

u in N such that

p[u’v’] = [uv].

PROOF. Let u’ be an element in N’ and U the

cyclic subgroup

of N such that

p[u’] = U.

Also let W be the

cyclic subgroup

^{of G such}

that

p [ u’v’ ] = W.

^Since

we have

[ U, W] = [U, V ] _ [W, V]

^{in G. Thus}

Proposition

¹ is ap-

plicable

^{to the situation at hand.}

By Proposition

^{1 the canonical}

homomorphism

^{cp from G onto}

G/N

maps W and V ^onto the same

subgroup

^{V of}

G/N.

the restriction of _{cp to} V is an

isomorphism

^{of V onto V. However it}

is also true

thatwnn = 1,

since V is an infinite

cyclic

group. Hence the restriction of cp to W is an

isomorphism

^{of W onto V. Therefore}

for any generator v of V there is a

unique

generator w ^{of W such} that Now if v is the generator ^{of V}

given

^{in the lemma and w is the} generator of ^W such that then w=uv where u is

unique

in U since N fl V =1.

Combining

^the

uniqueness

^{of w and the}

unique-

ness of u we have that there is a

unique u

^{in N such that}

p[u’v’] = [uv].

COROLLARY. If in Lemma 1 the

subgroup

^{N is in the center of}

G,

^then

[ u ] = U.

PROOF. This follows

immediately

^{from the}

corollary

^to

Proposi-

tion 1.

There is of course a ^« dual » result to Lemma 1 which asserts that under the conditions of Lemma 1 there is a

unique

^{element u in N}

such that

p[v’u’] = [vu].

^The

proof

^{uses the « dual » to}

Proposition

¹

which was mentioned after the

proof

^{of that}

Proposition.

^{It is} easy ^to

see that Lemma 1 and its dual show that for all u’ in

N’,

^{the conditions}

and

define functions on N’. The

corollary

^shows that if N is in the ^center of G these functions induce the

projectivity

^{which is the} restriction of p ^to N’.

We return once more to the

general

^{situation where G’ and G}

satisfy

^no

special

conditions. Let U be a

subgroup

^{of G.}

Suppose

^that

p and

(t’,

^{X G} induce the function

f(.; t’, t)

^{from U’} into U and that _p and

(v’,

^{X G} induce the function

f(.; v’, v)

^{from U’ into} U. _{It may}

happen (and frequently does)

^that

f(.; t’, t) = f (.; v’, v).

DEFINITION. Let 0 be a subset of G’ X G.

Suppose

^that every

pair (t’, (together

^with

p)

^{induces a function}

f(.; t’, t) (f(t’, t; .))

from U’ into U and that all of the _maps

f(.; t’, t) (f(t’,

^t;

^.))

^with

(t’,

^{are the same function}

f

^{from U’ into U. Then}

f

^{is said to be}

the function from U’ into U induced

by

^the

projectivity

p and the set 0 with 0 on the

right ^(left).

We write

f = f (.; 0) (or f=f(8; .))

^{if and}

only if f

^{is induced}

by

p and the set O with (8) on the

right (left).

4. An Extension of a

Projectively

Induced Function.

Throughout

this section let _p be a

projectivity

^{from some} group G’ onto a group

G,

let G be the

quotient

^{of G with} respect ^{to a normal}

subgroup ^N,

^{and let o} be the canonical

homomorphism

from G onto

G = G/N.

^In

studying subgroups

^{of the three} groups

G’, G,

^and

G,

^we shall use the

notation, H’, H,

^{and H for}

subgroups

^of

G’, G,

^and

G, respectively,

^{which are related}

by pH’ = H

^and

cpH = H.

LEMMA 2. Let N be a normal

subgroup

^of

^G,

^{V a} non-trivial

cyclic subgroup

^{of G such that}

^{V fl N=1,}

^{and U a} non-trivial

cyclic subgroup

of G such that

[ U, V I = U*V

^{in G. Let u’} and v’ be _genera- tors of U’ and

V’, respectively.

Then there is a

unique

^{element u in}

[ U, N]

^{and a}

unique

^{element v} in V such that

p [ u’v’ ]

⁼

[ uv ] .

^Moreover

PROOF. Let

p [ u’v’ ] = W.

Then from the

equations

which hold in

G’,

^{it follows that}

and

By Proposition

^{2 there are}

unique elements u, v, "iV,

in G such that

U= [u], V=[i;], W=[-Wl,

^and

^w=uv.

^Since

V fl N=1,

cp maps V

isomorphically

^onto V. It is also true that _{cp maps} W

isomorphically

onto

W,

^{since W is an infinite}

cyclic

group. Now take elements w E W and so that

cp(w)=w

^and

Let Then

cp(u)=wv-1=u,

^{and u is an element of}

[ U, N] . Obviously

^{w = uv and}

W = [ uv ] .

^{It is also} easy ^{to see} that there are no

elements

uoe[U, N]

^and such that For otherwise

[ U, V]

^{is not a free}

product

^without

amalgamation.

^{Thus w is the}

only

generator ^{of W which can} be written as a

product

^{of an element in}

[ U, N]

^{and an} element in V.

From V n [U, N ] =1

^{it follows that u}

and v are the

only

elements in

[ U, N]

^and

V, respectively,

^{such that}

w = uv. Theref ore u and v are the

unique

elements of

[ U, N]

^and

V,

respectively,

^{such that}

p [ u’v’ ] _ [ uv ] .

_

Since _cp maps V

isomorphically

^{onto V and since}

v=cp(v)

gene- rates

V,

^{it is} easy ^{to see} that v generates ^V. That is

[ v ] = V.

^Thus

[ U, V ] _ [ W, V ] _ [ uv, v] = [u, v ] .

^{Now let uo be a} generator of ^U and let be the

corresponding

generator ^{of U.} Since u is also

a generator ^of

U,

^{there is an}

integer k

^{such that}

Thus there is an element n in N such that Uo = ukn.

Hence

As we have

already

^{seen u is an} element of

[ U, N]

^{so that}

Thus

COROLLARY. If in Lemma 2 it is also assumed that N is in the center of

G,

^{then also}

[ u ] = U.

PROOF. Let uo be a

generator

^{of U.} From Lemma 2 we have that

Thus

where k is some

integer

^{and n is an element of}

[ u, v ]

n N. Since N is in the center of

G,

Therefore

In a similar _way it can be shown that Hence

[ u ] = U.

As in the case of Lemma 1 there is a dual result to Lemma 2 which

asserts that under the conditions of Lemma 2 there are

unique

^{u and}

v

(not necessarily

^{the same elements as in Lemma}

2)

ⁱⁿ

[U, N]

^and

V, respectively,

^{such that}

As in Lemma 2 these elements u and v also

satisfy

the condition

[v] = V, [u, v] = [U, V],

^and

[u, N] = [U, N].

If in Lemma 2 we had also assumed that V had infinite

order,

then the

pair (v’, v)

determined in the Lemma and the

projectivity

p would induce a function

f(.; v’, v)

^{from N’} into N. We would like ^to extend this function from N’ to all of

[ U’, N’].

^More

specifically

^we

would like ^to have our extended function be the

relation f

^defined

by f(u’)=u

^{if and}

only

^if

p[u’v’] = [uv]

for all u’ in

[ U’, N’]. However,

there _may be a u’ in U’ - N’ such that

for some u in

[ U, N].

^{In that case we have}

by

^{Lemma 2} that there is no u in

[ U, N ]

such that

p [ u’v’ ] _ [ uv ] ,

and the relation

f

^{is not}

a function from

[ U’, N’]

^into

[U, N] .

^In the remainder of this section

we show how to

pick

^{v’ and v so} that this

difficulty

^{can be avoided.}

LEMMA 3.

(Sadovsky).

a)

^{If u’ and v’ are} non-trivial elements of G’ such that

p[u’, v’ ]

=p[u’]*p[v’],

^{then there is a}

unique

^{element u in}

p[u’]

^{and a}

unique

^{element v in}

p [ v’ ]

such that

p[u’v’] = [uv].

The elements u and v

generate p[u’]

^and

p[v’], respectively.

b)

^{If in} part

a)

it is also assumed that

p[u’]

^and

p[v’]

^{are infinite}

cyclic

^or

equivalently

^that

p[u’, v’]

^{is a}

non-cyclic

^free group, then

p[(u’)m(v’)n] = [umvn]

^{for all} m, n= ± 1.

PROOF.

a)

^{In Lemma}

2,

let N =1. Given u’ and

v’,

^take

U = p [ u’ ]

^and

V=p[v’].

^{Then the}

hypotheses

^{of the lemma are}

fulfilled,

^{and the}

conclusion follows.

Hence

W = [ uv -1 ] .

^{The other} cases, i.e.

m = -1,

n=+l and m,

n = -1,

^{can be} verified in a similar _way.

In Lemma 4 we use two facts from _group

theory

^without

proof.

First if

G = [ ui , u2 ] * [ t ] * [ v ]

^where ui , u2 , t, v ^are non-trivial elements of G with ul , u2 not

necessarily distinct,

^then

H = [ uit, U2V]

^{is a non-}

cyclic

^free

subgroup

of G. Second let

G = [ a, b]

^{and let}

G = [ a, ~ ]

^be

the

image

of G under some

homomorphism

^{cp where}

a=cp(a)

^and

If G is a

non-cyclic

^free group then G is also a

non-cyclic

free _group.

LEMMA 4. Let N be a normal

subgroup

^of

^G,

^{V a} non-trivial

subgroup

of G such that

V f 1 N =1,

^{and U a} non-trivial

subgroup

^{of G}

such that

[ U, V ] = U*V.

^{Let t’ and v’ be two} elements of G’ such that

p[t’, v’]

^{is a}

non-cyclic

^free

subgroup

of V. Let u’ be an element of

[ U’, N’ ] .

Then there are

unique

^{u in}

[ U, N ] , unique t

ⁱⁿ

p[t’],

^and

unique v

ⁱⁿ

p [ v’ ]

^{such that}

PROOF. First let u’ be an element of the

complement

^{of N’ in}

[ U’, N’ ] . By

^{Lemma 2 we} know there ^are

unique

^{ui in}

[p[u’], N]

and t in

p[t’]

^{such that}

Similarly

^{there are}

unique

^u2

in

[p[u’], N],

^{and v in}

p[v’]

such that

p [ u’v’ ] _ [ u2v ] .

^{We shall}

show that and that

p[t’v’] = [tv].

The _group

[t, v]

^{is a free}

subgroup

of V since

p[t’, v’] _ [ t, v].

Because V fl N=1 the canonical

homomorphism

^{cp from G onto G maps}

[t, v ] isomorphically

^onto

[1, v ] , where = cpt, T=gpv.

^Hence

[ t, v ]

is a free

subgroup

of V. From this fact and from the fact that

[U, V ] = U*V

^{in G it follows that is a free}

subgroup

^of

^G,

where and

u2 = cpu2 .

^{Bu t}

[ ult, is

the

image

^of

[ ult,

under the canonical

homomorphism

^{cp. Therefore is a free}

subgroup

^{of G.}

We are now able to

apply

^{Lemma 3 b to the}

cyclic

group

generated by ( u’~’) - lu’v’ _ ( t’) - lv’.

Thus there are

integers k,

^{l = ± 1 such that}

and

integers

m, n = ± 1 such that

But then

(t-1uï1 1)’(u2v)l

^and

(t-1),Vn

^must both generate ^{the same infinite}

cyclic

group in G. Therefore there is an

integer /=±1

^{such that}

Mapping

^{both sides of this}

equation by

^{cp into G we have}

By Proposition

³

Therefore in

G,

^and

consequently

^{in G. Also}

since _m,

n= -E-1, p[(t’)-Iv’l =

^and

by

^{Lemma 3 b}

p[t’v’] = [tv].

Now let u’ be an element of N’.

By

^{Lemma 3 b}

pick

^{elements t}

and v in G so that

p[t’] _ [t], p[v’] _ [v],

^and

p[t’v’]=[tv]. By

Lemma 1 there are

unique

^{ul and u2 in N such that}

We shall show that ul = u2 .

Again [ t, v ]

^and

[ t, v ]

^are

non-cyclic subgroups

^{of V and}

^V, respectively.

^{Since and we} have that

[ult, u2v]

is a

non-cyclic

^free

subgroup

^{of G. As in the}

previous

^{case we}

apply

Lemma 3 b to the

cyclic subgroup generated by

Thus there are

integers k,

^{l = ± 1 such that}

Since t and v were

already

^{selected so that}

p[t’v’] _ [tv],

^{we have}

by

^{Lemma 3 b that}

Thus there is an

integer /=±1

^{such that}

in G.

Mapping

both sides of this

equation by

cp into G we have

Since 1 and v are a system ^{of free} generators ^{for a free} group, ^we have that

Hence and ui = u2 .

In addition to

being unique

ⁱⁿ

[p[u’], N]

the element u is

unique

in all of

[ U, N].

^That

is,

there is no element ui in

[U, N]

^{such that}

N]

^{and For if there were} the elements

u = cp(u)

^and

v = cp(v)

would have to

satisfy

^{the conditions}

and

[U-V]=[-ui-vl

^{in U*V.} But this is

clearly impossible

^{in a free}

product

^without

amalgamation.

Thus we see that for each element u’ in

[ U’, N’]

^{there is a}

unique

element u in

[ U, N ]

such that

p[u’v’] = [uv]. Therefore,

^{for all u’}

in

[ U’, N’],

the condition

defines a function

f(.; v’, v)

^from

[U’, N’]

^into

[U, N].

^{In a similar}

way it can be shown that there exist

analogous

^functions

f(.; t’, t), f ( v’,

^v;

.)

^and

f(t’, t; .)

^from

[ U’, N’]

^into

[ U, N].

From Lemma 4 and its ^« dual » it follows that

and that

In the next section we show that all four functions are

equal.

^{It is}

also shown there that the function induced

by

^the

projectively

p and the

pairs (t’, t)

^and

(v’, v)

^{is an}

isomorphism

^from

[ U’, N’]

^onto

[ U, N].

5.

Isomorphisms

^Induced

by

^a

Projectivity.

Throughout

this section let _p be a

projectivity

^{from some} group G’ onto a group

G,

^{N a normal}

subgroup

^of

^G,

^{and cp the canonical}

homomorphism

^{from G onto}

G = G/N.

^{As in Section 4 if H is a sub-}

group of G we write H_’ for the

subgroup

of G’ such that

pH’ = H and

H for the

subgroup

^{of G such that}

^cpH

^{= H.}

LEMMA 5.

Suppose

^{there is a}

non-cyclic

^free

subgroup

^{V of G}

such that V f 1 N =1. Then N’ is a normal

subgroup

^of

[ N’, V’].

PROOF. Let u’ and v’ be elements of N’ and

V’, respectively.

^We

wish to show that v’-lu’v’ is an element of N’. Let t’ be one of the

numerous elements of V’ such that

p[t’, v’ ]

^{is a}

non-cyclic

^{free sub-}

group of V. For

such t’, u’,

^{and v’ we shall show that the}

subgroup H’ _ [ t’u’v’, t’v’]

^meets

[ v’, N’]

ⁱⁿ

N’,

^{that is}

But

(v’)-lu’v’=(t’v’)-1(t’u’v’)

^{is an} element of the intersection on the

left,

^and

consequently

^an element of N’.

By

^{Lemma 3 there are}

unique

elements t and v in G such that

p[t’]=[t], p[v’] _ [v]

^and

p[t’v’] _ [tv]. By

^{Lemma 1 there is a}

unique

^{u in N} such that

p [ t’u’ ] _ [ tu ] .

^{We let}

H=pH’, H=cpH, t = cpt,

^{and We} first show that H n ^or H n must be the

identity subgroup

^{of G. Then we show that}

H=[Fvl

^{and that}

From this we conclude that both

N’ ]

^and

JFff1[Y, N’]

^are

contained in N’.

_

Since

[ tu, v]

and

[ t, v ]

^{have the same}

image

^under cp in

G,

^it follows that

[ tu, v ]

^{is a}

non-cyclic

^free

subgroup

^{of V.}

By

^{Lemma 3}

there are

integers k,

^{t = ± 1 such that Thus}

tv]

^{and The four}

possibilities

^{f or H are}

It is _easy to see that

Hence H f 1 = 1 or H f 1

[ v ]

^=1.

Consequently ^N]

^{or H f}

1 [ v, N]

is contained in N. But this means that

N’]

^or

H’ n [v’, N’]

^is

contained in N’.

Up

^{to this}

point

^{we have shown that I or}

(v’) -’u’v’

^{is an}

element of N’. It remains to be shown that both elements are in N’.

We _suppose without loss of

generality

^{that in N’. Then}

t’u’v’=u’it’v’. By

^{Lemma 1 there is a}

unique

^{element u, in N such that}

Hence and

But this means that both

H’ n [t’, N’ ]

^and

H’ n [v’, N’ ]

^{are in N’.}

Hence

(v’)-lu’v’

^{must be an element of}

N’,

^{for all v’ in V’.}

COROLLARY. With the same notation as in the

proof

^{of Lemma}

5 we have that for each element u’ in N’ there is a

unique

^{element u}

in N such that

p[t’u’]

^.-

[tu], p[u’v’] _ [uv],

^and

p[t’u’v’l = [tuv].

PROOF. Let u’ be some element of N’. We have

already

^{seen that}

there are

unique

^{elements u, and u2} in N such that and

p[x/T/]=[M2~]. By

^{Lemmas 5 and 1 there is a}

unique

^{element u3 in}

N such that